The molar conductance of a 0.01 M solution of acetic acid was found to be 16.30 Ω^-1 cm^2 mol^-1 at — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Formula for Degree of Dissociation ($\alpha$) The degree of dissociation ($\alpha$) of a weak electrolyte represents the fraction of solute molecules that ionize in a solution at a specific concentration. It is mathematically defined as the ratio of the molar conductivity at concentration $C$ ($\Lambda_m^C$) to the limiting molar conductivity at infinite dilution ($\Lambda_m^\circ$): $$\alpha = \frac{\Lambda_m^C}{\Lambda_m^\circ}$$ The percentage dissociation represents the percentage of molecules that dissociate in the solution and is given by: $$\text{Percentage dissociation} = \alpha \times 100\%$$ We are given the molar conductivity of a $0.01\text{ M}$ acetic acid solution: $$\Lambda_m^C = 16.30\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$$ To find the percentage dissociation, we must first calculate the limiting molar conductivity ($\Lambda_m^\circ$) of acetic acid ($\ce{CH3COOH}$). Step 2 - Apply Kohlrausch's Law of Independent Migration of Ions According to Kohlrausch's Law, the limiting molar conductivity of an electrolyte at infinite dilution is equal to the sum of the individual limiting ionic conductivities of its constituent cations and anions. Acetic acid dissociate as follows: $$\ce{CH3COOH(aq) <=> CH3COO^-(aq) + H^+(aq)}$$ Therefore, the limiting molar conductivity of acetic acid is: $$\Lambda_m^\circ(\ce{CH3COOH}) = \lambda^\circ(\ce{H^+}) + \lambda^\circ(\ce{CH3COO^-})$$ Given: * $\lambda^\circ(\ce{H^+}) = 349.8\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$ * $\lambda^\circ(\ce{CH3COO^-}) = 40.9\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$ Substitute these values into the Kohlrausch equation: $$\Lambda_m^\circ(\ce{CH3COOH}) = 349.8\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1} + 40.9\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$$ $$\Lambda_m^\circ(\ce{CH3COOH}) = 390.7\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$$ Step 3 - Calculate the Degree of Dissociation ($\alpha$) and Percentage Dissociation Substitute the value of $\Lambda_m^C$ and the calculated $\Lambda_m^\circ(\ce{CH3COOH})$ into the formula for the degree of dissociation: $$\alpha = \frac{\Lambda_m^C}{\Lambda_m^\circ}$$ $$\alpha = \frac{16.30\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}}{390.7\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}}$$ $$\alpha \approx 0.04172$$ Now, calculate the percentage dissociation: $$\text{Percentage dissociation} = \alpha \times 100\%$$ $$\text{Percentage dissociation} = 0.04172 \times 100\%$$ $$\text{Percentage dissociation} = \mathbf{4.172\%}$$ Thus, the percentage of acetic acid dissociated at a concentration of $0.01\text{ M}$ is exactly $4.172\%$. Step 4 - Evaluate and Explain the Options * **Option (A) is incorrect:** The value $0.04172\%$ represents the fractional degree of dissociation ($\alpha$) itself, without multiplying by $100$ to convert it to a percentage. * **Option (B) is correct:** As mathematically shown, dividing the molar conductivity by the limiting molar conductivity and converting to a percentage yields $4.172\%$. * **Option (C) is incorrect:** This value ($41.72\%$) represents a decimal shift error by a factor of $10$, which results from an arithmetic mistake during division. * **Option (D) is incorrect:** This value ($0.4172\%$) also represents a decimal placement error of $10$, which does not correspond to the correct calculations. $$\text{Correct Option: } \boxed{\text{B}}$$